I have a datastore entity called Game and two fields in it called playerOne and playerTwo. Either of these fields stores a username.
I need to search on the Game entity and return a MAX of 30 games where the username can be either playerOne OR playerTwo...
So in a relational database you would go:
SELECT * FROM Game WHERE playerOne='username' OR playerTwo='username' LIMIT 30
But in big table you can't filter on more than one field! I can't fetch 10 from one and 10 from the other as the number from each can be variable and in createdDate order.
How would you do this in your datastore?
The quick answer is create a StringListProperty that contains [player_a, player_b] and then simply use the multi-value index made out of that:
games = Game.all().filter("players =", player_find)
You can not do an OR query on the datastore using different fields. If you have to keep your current entity model then you have to do two queries.
1) filtering on playerOne and limiting to 30
2) filtering on playerTwo and limiting to (30 - result size of query one)
Then merge the results in memory to produce the final set of 30.
Now if you also want some ordering by date, then it will get more tricky. However the SQL query you wrote doesn't have any ordering so I omitted it aswell.
However if you can change the entity model then a good way to achive what you want is to have a single field containing a list of both usernames.
Then you can do a simple query in the style of:
SELECT * FROM Game WHERE playerBoth = 'username'
Related
I have a Kind Students which stores the details of favorite colors of all students. They are allowed to pick their favorite color from a set of three colors : {Red,Blue,Green}
Let us assume there are 100 students, my code is like this for every student :
Entity arya = new Entity("Student","Arya");
arya.setProperty("Color","Red");
Entity robb = new Entity("Student","Robb");
robb.setProperty("Color","Green");
..
..
Entity jon = new Entity("Student","Jon");
jon.setProperty("Color","Blue");
How to find out how many students liked a particular color(say Red) in this Student Kind ? What Query I should write to fetch the count ?
Thanks in advance
The number you seek would be the number of items in the result of a query with an equality filter on the Color property.
You could use a keys-only query (a special kind of projection query) for this purpose, faster and less expensive:
Keys-only queries
A keys-only query (which is a type of projection query) returns just
the keys of the result entities instead of the entities themselves, at
lower latency and cost than retrieving entire entities.
...
A keys-only query is a small operation and counts as only a single
entity read for the query itself.
Something along these lines (but note that I'm not a java user, the snippet is based only on the documentation examples)
Query<Key> query = Query.newKeyQueryBuilder()
.setKind("Student")
.setFilter(PropertyFilter.eq("Color", "Red")
.build();
I agree with the Dan Cornilescu's answer. Here is a direct Datastore API usage. I have prepared the request body for your use-case. You can run it by just adding your Project Id. This will return the entities that matches with the filter then you can count the number of them.
I am working currently on telecom analytics project and newbie in query optimisation. To show result in browser it takes a full minute while just 45,000 records are to be accessed. Could you please suggest on ways to reduce time for showing results.
I wrote following query to find call-duration of a person of age-group:
sigma=0
popn=len(Demo.objects.filter(age_group=age))
card_list=[Demo.objects.filter(age_group=age)[i].card_no
for i in range(popn)]
for card in card_list:
dic=Fact_table.objects.filter(card_no=card.aggregate(Sum('duration'))
sigma+=dic['duration__sum']
avgDur=sigma/popn
Above code is within for loop to iterate over age-groups.
Model is as follows:
class Demo(models.Model):
card_no=models.CharField(max_length=20,primary_key=True)
gender=models.IntegerField()
age=models.IntegerField()
age_group=models.IntegerField()
class Fact_table(models.Model):
pri_key=models.BigIntegerField(primary_key=True)
card_no=models.CharField(max_length=20)
duration=models.IntegerField()
time_8bit=models.CharField(max_length=8)
time_of_day=models.IntegerField()
isBusinessHr=models.IntegerField()
Day_of_week=models.IntegerField()
Day=models.IntegerField()
Thanks
Try that:
sigma=0
demo_by_age = Demo.objects.filter(age_group=age);
popn=demo_by_age.count() #One
card_list = demo_by_age.values_list('card_no', flat=True) # Two
dic = Fact_table.objects.filter(card_no__in=card_list).aggregate(Sum('duration') #Three
sigma = dic['duration__sum']
avgDur=sigma/popn
A statement like card_list=[Demo.objects.filter(age_group=age)[i].card_no for i in range(popn)] will generate popn seperate queries and database hits. The query in the for-loop will also hit the database popn times. As a general rule, you should try to minimize the amount of queries you use, and you should only select the records you need.
With a few adjustments to your code this can be done in just one query.
There's generally no need to manually specify a primary_key, and in all but some very specific cases it's even better not to define any. Django automatically adds an indexed, auto-incremental primary key field. If you need the card_no field as a unique field, and you need to find rows based on this field, use this:
class Demo(models.Model):
card_no = models.SlugField(max_length=20, unique=True)
...
SlugField automatically adds a database index to the column, essentially making selections by this field as fast as when it is a primary key. This still allows other ways to access the table, e.g. foreign keys (as I'll explain in my next point), to use the (slightly) faster integer field specified by Django, and will ease the use of the model in Django.
If you need to relate an object to an object in another table, use models.ForeignKey. Django gives you a whole set of new functionality that not only makes it easier to use the models, it also makes a lot of queries faster by using JOIN clauses in the SQL query. So for you example:
class Fact_table(models.Model):
card = models.ForeignKey(Demo, related_name='facts')
...
The related_name fields allows you to access all Fact_table objects related to a Demo instance by using instance.facts in Django. (See https://docs.djangoproject.com/en/dev/ref/models/fields/#module-django.db.models.fields.related)
With these two changes, your query (including the loop over the different age_groups) can be changed into a blazing-fast one-hit query giving you the average duration of calls made by each age_group:
age_groups = Demo.objects.values('age_group').annotate(duration_avg=Avg('facts__duration'))
for group in age_groups:
print "Age group: %s - Average duration: %s" % group['age_group'], group['duration_avg']
.values('age_group') selects just the age_group field from the Demo's database table. .annotate(duration_avg=Avg('facts__duration')) takes every unique result from values (thus each unique age_group), and for each unique result will fetch all Fact_table objects related to any Demo object within that age_group, and calculate the average of all the duration fields - all in a single query.
I have a following model in the GAE app.
class User
school_name = db.StringProperty(Indexed=True)
country = db.StringProperty(Indexed=True)
city = db.StringProperty(Indexed=True)
sex = db.StringProperty(Indexed=True)
profession = db.StringProperty(Indexed=True)
joined_date = db.DateTimeProperty(Indexed=True)
And I want to filter the users by combinations of these fields. Result of the filter should show a user at first who is joined recently. So which means any query end by order operation, I suppose. like that:
User.all().filter('country =','US').filter('profession =','SE').order('-joined_date')
User.all().filter('school_name =','AAA').filter('profession =','SE').order('-joined_date')
....
User.all().filter('sex =','Female').filter('profession =','HR').order('-joined_date')
All these fields combination would be C(5,1)+C(5,2)+...+C(5,5) = 31.
My question is to implement it, do I need to create indexes for all these cases(31) in the Google AppEngine. Or can you suggest other way to implement it?
Note: C(n,k) is combination formula, see more on http://en.wikipedia.org/wiki/Combination
Thanks in advance!
You have several options:
Create all 31 indexes, as you suggest.
Do the sorting in memory. Without a sort order, all your queries can be executed with the built-in merge-join strategy, and so you won't need any indexes at all.
Restrict queries to those that are more likely, or those that eliminate most of the non-matching results, and perform additional filtering in memory.
Put all your data in a ListProperty for indexing as "key:value" strings, and filter only on that. You will need to create multiple indexes with different occurrence counts on that field (eg, indexing it once, twice, etc), and it will result in the same number of index entries, but fewer custom indexes used.
How would one go about writing a query that selects items 2000-2010 out of a collection of 10000 objects in the data store.
I know that it can be done like this in GQL:
select * from MyObject limit 10 offset 2000
According to the documentation, when using an offset the engine will still fetch all the rows, only not return them, thus making the query perform in a way that corresponds linearly with the value of offset.
Is there any better way? Such as using a pseudo ROWNUM column like one could do in other types of data stores.
There's no way to efficiently page using offsets, except to cache the results. You can, however, use datastore cursors to implement paging using a 'bookmark' type approach.
Besides using cursors you can also use a sort order approach. For example:
SELECT * FROM MyObject ORDER BY field LIMIT 10;
for the first 10 objects and then for the next 10 objects, etc.
SELECT * FROM MyObject WHERE field > largestFieldValueFromPreviousResult ORDER BY field LIMIT 10;
Field could even be a key if you don't have another appropriate field. Here is a more complete example:
http://code.google.com/appengine/articles/paging.html
Select all records, ID which is not in the list
How to make like :
query = Story.all()
query.filter('ID **NOT IN** =', [100,200,..,..])
There's no way to do this efficiently in App Engine. You should simply select everything without that filter, and filter out any matching entities in your code.
This is now supported via GQL query
The 'IN' and '!=' operators in the Python runtime are actually
implemented in the SDK and translate to multiple queries 'under the
hood'.
For example, the query "SELECT * FROM People WHERE name IN ('Bob',
'Jane')" gets translated into two queries, equivalent to running
"SELECT * FROM People WHERE name = 'Bob'" and "SELECT * FROM People
WHERE name = 'Jane'" and merging the results. Combining multiple
disjunctions multiplies the number of queries needed, so the query
"SELECT * FROM People WHERE name IN ('Bob', 'Jane') AND age != 25"
generates a total of four queries, for each of the possible conditions
(age less than or greater than 25, and name is 'Bob' or 'Jane'), then
merges them together into a single result set.
source: appengine blog
This is an old question, so I'm not sure if the ID is a non-key property. But in order to answer this:
query = Story.all()
query.filter('ID **NOT IN** =', [100,200,..,..])
...With ndb models, you can definitely query for items that are in a list. For example, see the docs here for IN and !=. Here's how to filter as the OP requested:
query = Story.filter(Story.id.IN([100,200,..,..])
We can even query for items that in a list of repeated keys:
def all(user_id):
# See if my user_id is associated with any Group.
groups_belonged_to = Group.query().filter(user_id == Group.members)
print [group.to_dict() for group in belong_to]
Some caveats:
There's docs out there that mention that in order to perform these types of queries, Datastore performs multiple queries behind the scenes, which (1) might take a while to execute, (2) take longer if you searching in repeated properties, and (3) will up your costs with more operations.